Area inside Polar Curves

  • Thread starter Gibybo
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[SOLVED] Area inside Polar Curves

Homework Statement



I have spent several hours beating myself up over this and I just can't seem to solve it. It's the only problem I haven't gotten correct and it is particularly frustrating. Can you math gods here save me? :)

Find the area of the region that lies inside both curves
[tex]r = 6 \sin (2 \theta) , \quad r = 6 \sin (\theta) [/tex]

Homework Equations



Area inside a single polar curve: [tex]\frac{1}{2}\int{f(\theta)^{2} d\theta}[/tex]

The Attempt at a Solution



They intersect at [tex]\frac{\pi}{3}[/tex] and 0
So, I take the area of the second equation from 0 to pi/3, and the area of the first equation from pi/3 to pi/2 since pi/2 is where it meets back at the origin.

[tex]\frac{1}{2}\int_{0}^{\pi/3}{(6\sin(\theta))^{2} d\theta} + \frac{1}{2}\int_{pi/3}^{\pi/2}{(6\sin(2\theta))^{2} d\theta} = \frac{-9(3\sqrt(3)-4\pi)}{8} \approx 8.2915[/tex]
The website (WebWork) that we input the answer to says it is not correct though, and I cannot figure out any other way that would make sense.
 

Answers and Replies

  • #2
HallsofIvy
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Why are you adding the two integrals? The area between [itex]r= f(\theta)[/itex] and [itex]r= g(\theta)[/itex] is the integral of [itex]|f(\theta)- g(\theta)|[/itex].
 
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  • #3
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Why are you adding the two integrals? The area between [itex]r= f(\theta)[/itex] and [itex]r= g(\theta)[/itex] is the integral of [itex]|f(\theta)- g(\theta)|.

Actually, I believe that his integrals may be correct. Whether or not the actual underlying integration is correct I didn't check. When two polar curves are used, the formula you gave is generally the correct approach -- but only when the resulting region is topologically equivalent to an annulus (i.e. one curve resides inside the other curve within the given domain). However, the radius from 0 to pi/3 is dependent only on 6sin(theta), whereas from pi/3 to pi/2, the radius is solely dependent on 6sin(2theta). In this case, a graph and some shading will definitely clarify the need for two integrals.

What needs to be fixed, however, is that the sum of the two integrals above only gives 1/2 of the area contained within both curves. 6sin(2theta) is essentially a "rose petal" curve with four "petals," two of which intersect 6sin(theta). Assuming your math is correct, you should be able to just double the value to get the correct area contained within both curves.
 
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  • #4
Avodyne
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This looks right to me, but it covers the first quadrant only. Maybe you should multiply by 4?
 
  • #5
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Best I could do was a MS Paint recreation:
http://www.warcraftcheese.com/math/polar%20area.jpg [Broken]
Note that the loop on the right should be pedal shaped, does not dip below the x-axis on the interval [0,pi/2], and is 6sin(2t).

The question says inside both, so the idea is to calculate the inner area of the bottom-right side of that region (from origin to intersection point) using 6sin(t), and add it to the upper-left area using 6sin(2t).

EDIT:
Assuming your math is correct, you should be able to just double the value to get the correct area contained within both curves.
This looks right to me, but it covers the first quadrant only.
Oh that's right, thanks so much :)
 
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